a: \(P=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

b: \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\)

Khi x=4-2căn 3 thì \(P=\dfrac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\dfrac{3}{\sqrt{3}-1}=\dfrac{3\sqrt{3}+3}{2}\)

a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

Ta có:

\(x^2+1=x^2+xy+yz+zx\)

           \(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

\(=y\left(x+z\right)+z\left(x+z\right)\)

\(=\left(x+z\right)\left(y+z\right)\)

CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)

Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)

\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)

 Tương tự như thế, ta được

\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{2}+1-\sqrt{2}+1\)

=2

Thay x=2 vào A, ta được:

\(A=\dfrac{-3}{3+\sqrt{2}}=\dfrac{-9+3\sqrt{2}}{7}\)

\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)

Ta có :A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\) -\(\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\) 

=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)-2

=\(\dfrac{-\sqrt{x}}{\sqrt{x}+1}\)

thay vào A=\(\dfrac{-2}{3}\)

b)

A=-1+\(\dfrac{1}{\sqrt{x}+1}\) \(\ge\) -1+\(\dfrac{1}{1}\)=1(vì \(\sqrt{x}\)\(\ge\) 0)

Dấu bằng xẩy ra\(\Leftrightarrow\) x=0

chỗ đó cho thêm x-1 nha

đấu >= thay thành <= rùi nhân thêm x-1>=-1 nữa là lớn nhất bằng 0

a) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)

b) \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)  (*)

Thay (*) vào B , ta được : \(B=\dfrac{2-\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{-\sqrt{3}+1}{3\sqrt{3}+3}\)

Bạn santa làm sai r nha!

a, ĐKXĐ: x \(\ge\) 0; x \(\ne\) 4; x \(\ne\) 0

B = \(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)

B = \(\left(\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right)\)

B = \(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

B = \(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3}\)

B = \(\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{3\sqrt{x}\left(\sqrt{x}+1\right)}\)

B = \(\dfrac{2-\sqrt{x}}{3\sqrt{x}}\) (Đoạn này bạn kia viết sai đề mà vẫn đúng kết quả được?)

Vậy ...

b, Ta có: x = 4 + 2\(\sqrt{3}\) = (\(\sqrt{3}\) + 1)² (TMĐK)

\(\Rightarrow\) \(\sqrt{x}\) = \(\sqrt{3}+1\) (1)

Thay (1) vào B ta được:

B = \(\dfrac{2-\sqrt{3}-1}{3\left(\sqrt{3}-1\right)}\) = \(\dfrac{1-\sqrt{3}}{-3\left(1-\sqrt{3}\right)}\) = \(\dfrac{-1}{3}\)

Vậy ...

Chúc bn học tốt!

mình làm lại nhé :

đkxđ : \(\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{3}\)

\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)

câu b làm như kia là oke rồi nhé <3